Chapter 6 Shear Strength of Soil Mohr-Coulomb Failure ... Strength of Soil ... Shear Strength...

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- 1 - Chapter 6 Shear Strength of Soil Mohr-Coulomb Failure Theory Shear Strength Parameters Determination of Shear Strength Parameters Direct shear test Unconfined compression test UU triaxial test Vane shear test Pore Pressure Parameters

Transcript of Chapter 6 Shear Strength of Soil Mohr-Coulomb Failure ... Strength of Soil ... Shear Strength...

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Chapter 6

Shear Strength of Soil

Mohr-Coulomb Failure Theory

Shear Strength Parameters

Determination of Shear Strength Parameters

Direct shear test

Unconfined compression test

UU triaxial test

Vane shear test

Pore Pressure Parameters

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Chapter 6

SHEAR STRENGTH OF SOIL

6.1 Introduction

If the load or stress in a foundation or earth slope is increased to cause an unacceptably large

deformation, the soil in the foundation or slope is considered as failed against its strength.

Geotechnical engineers always consider strength of the soil as the shear strength. This is because

the tensile strength of soil is very small and practically accepted to be zero. The applied

compressive load eventually causes the soil to fail in shear in a rupture surface. As the soil is a

granular or particulate material the excessive deformation, as mentioned, generally occurs due to

sliding, rolling or rearrangements in some way of the particles with in a soil mass or simply due to

shearing. This phenomenon is illustrated in Fig. 6.1.

Fig. 6.1: Illustration of soil failing in shear due to sliding and rolling of particles past each other.

Therefore, the sliding capability of a soil, while to support a loading from a structure or to

support its own overburden or to sustain a slope in equilibrium, is called shear strength of

soil. As it is referred to the strength, it is the maximum or ultimate shear stress that the soil can

sustain to continuous shear displacement of soil particles along a surface of rupture.

6.2 Importance of Shear Strength in Soil Mechanics

In many of the soil mechanics problems, the shear strength of the soil emerges as one of the most

important characteristics. The examples are

Foundation design

Earth and rockfill dam design

Highway and airfield design

Stability of slopes and cuts

Lateral earth pressure

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6.3 Contributing Factors of Shear Strength of Soil

The shear strength of the soil may be attributed to three basic components:

Resistance due to interlocking of the particles

Frictional resistance between the individual soil grains, which may be either of the

sliding or rolling frictions or the both.

Adhesion between soil particles called cohesion

First two sources are mainly responsible for the shear strength of granular soils; second and

third for the cohesive soils. Whereas, third one is only responsible for shear strength of highly

plastic clays. It is neither easy nor practical to clearly distinguish the effects of these components

on the shear strength of the soil, because these components in turn are influenced by many

factors like:

(i) Heterogeneous nature of soil that typify most soil masses

(ii) The water table location and moisture contents

(iii) The drainage facility involving pore water pressure

(iv) The type and nature of construction

(v) Stress history

(vi) Structural disturbance of soil

(vii) Chemical action

(viii) Environmental conditions

Therefore, the shear strength of a soil cannot be interpreted in a very simple way since a soil can

significantly exhibit different shear strength at various field conditions. The shear strength (or its

parameters) and deformation of a foundation can be obtained either from laboratory tests on

carefully extracted field samples or from in situ tests. However, the understanding of these

interrelated and complex matters needed to have a clear in view of the following basic aspects.

Friction between solid bodies

Cohesion between the soil particles

Stresses on the planes passing through a point represented by Mohr Circle

diagram

6.3.1 Friction between solid bodies

This is similar to classic sliding friction problem from basic physics or mechanics. The force that

resists sliding equals to normal force multiplied by a coefficient of friction, ..

Let us consider a block of weight, W placed on a horizontal plane having frictional interface as

shown in Fig. 6.2. The forces acting on the block, Fig. 6.2a, are gravity and the reaction of the

plane. Although a friction resistance is available, it does not come into play since there is no

horizontal force applied to the block.

Now, a horizontal force P1 is applied to the block, Fig. 6.2b. If P1 is small, the block will not move;

the resultant Ra makes an angle commonly known as angle of obliquity of force P1. For

equllibrium, a resisting horizontal force F1 must therefore exist to balance P1. The angle is less

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than the maximum angle available, for the maximum horizontal force, known as angle of friction, .

In Figure, N is the component of the weight normal to the plane.

If the applied horizintal force is increased to P2 , as shown in Fig 6.2c, the friction force also

increases to a value of F2. It reaches to a maximum value when the angle of obliquity is equal to

the angle of friction, . At this point a state of impending motion exists. For the range of value 0 <

< no slip or sliding occurs. At this point of impending motion, the maximum available frictional

force F can be related to the normal resisting force by a coefficient as given by

NF (6.1)

Fig. 6.2 Illustration of frictional forces. (a) Development of no friction; (b) Development of

friction; (c) Friction impending motion; (d) Friction with motion.

The coefficient of friction is independent of the area of contact. It is, however, strongly

dependent on the nature of the surface in contact, the type of material, the condition (wet or

dry) of the surface and so on.

If the horizontal force is increased to a value of P3 as shown in Fig. 6.2d, such that the angle is

greater than , the block will start sliding. The frictional force can not exceed the value given by F =

N, and therefore the block will accelerate in the direction of applied horizontal force. Fig. 6.2c

shows that = F/N = tan , which also means that / = tan .

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6.3.2 Concept of Cohesion

The soil mass consists of granular materials. There exists two forces namely, body force and

Van der Wall force, between two bodies which usually govern the the properties of granular

mass. The body force is usually referred to as gravitational force where the mass of the bodies are

predominent and larger particles are influenced by this force. Whereas, Van der Wall force is due to

electrical surface charges and a bonding between the particles occurs in presence of water which

helps to generate various chemical bonding among the particles. This adhesion due to bonding is

known as cohesion for soil. In case of smaller soil particles the specific surface area (area per unit

mass) is significantly higher as compared to the coarse grained soil. As the electrical charges and

hence the Van der Wall force are directly proportional to surface charge, the property of fine grained

soil is governed by cohesion of the soil. The body force here is insignificant because of their smaller

mass weight. Similar thing happens to coarse grained soil ; the Van der Wall force here has a very

small influence.

However, this cohesion parameter is dependent on stress history that is mode of formation of soil

deposit. In soils like overconsolidated clays, partially saturated soils and cemented soils the

individual particles are bonded togather giving rise to cohesion. As such, this parameter can be

viewed as a source of shear strength that is independent of normal force.

6.3.3 Stresses on planes passing through a point and Mohr’s circle

According to principles of mechanics, both normal and shear stresses act on planes passing through

a point that has been subjected to external load. These stresses may be represented graphically by an

extremely useful device known as Mohr’s circle of stress. The device is based on a unique point on

a circle called the pole or origin of planes. This point has such a useful property that “ any straight

line drawn through this point (pole) will intersect the circle at a point which represents the state of

stress on a plane inclined at the same the orientation in the space of the line.” According to this Pole

method, atleast two lines from known points of stresses on Mohr‟s circle parallel to the

corresponding planes on which they act are drawn to intersect a point on the circle which is the pole.

Once the pole is known the stresses on any plane can be found by drawing a line from the pole

parallel to that plane; the coordinates of the point of intersection with the Mohr‟s circle determine

the stresses on that plane. Mohr‟s circle in soil mechanics follows certain sign convention.

The compressive stress is considered as positive

The counter-clockwise shear stress is positive

The horizontal and vertical coordinates represents normal and shear stresses respectively. For example, in Fig. 6.3(a) the normal and shear stresses are acting on the planes of a recangular element. Now, let us consider a plane on the element (Fig. 6.3a) making an angle θ with the horizontal. If a line is drawn parallel to this plane from the pole point PD to intersect the

circumference of the Mohr‟s circle at a point that will represent the stresses ( ’, ) acing on the plane in question. In most of the soil mechanics problems only the top half of the Mohr‟s circle is drawn for convenience. Mohr‟s circles are usually drawn for principal stresses and the pole lies on the normal

stress ( ’) axis. The other point is that there would be two failure planes in such a case. This has got a little significance because of the fact that in practical field condition the particulate material soil will initiate its failure at the weakest of the two planes.

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Mohr Circle of stress

Soil element

’1

’1

’3’3

222

22

'

3

'

1

'

3

'

1'

'

3

'

1

Cos

Sin

Resolving forces in and directions,

2'

3

'

1

2'

3

'

1'2

22

(a) Mohr Circle of stress

2'

3

'

1

2'

3

'

1'2

22’

2

'

3

'

1

2

'

3

'

1

'

3

'

1

PD = Pole w.r.t. plane

’,

(b)

Fig. 6.3 (a) Stresses on an element; (b) Illustration of Pole of Mohr circle.

6.4 Mohr-Coulomb Failure Theory

Mohr (1900) presented a theory for rupture of materials. This states that the failure in materials

occurs with a critical unique combination of normal and shear stresses on the rurture surface and not

solely by either of the maximum normal or maximum shear stresses. This can be expressed as a

function in the form

ff (6.2)

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The stress f at failure (rupture) plane is called the shear strength of the material. f is the normal

stress on the failure plane. Fig. 6.4 shows the relationship as expressed by Eq. (6.2) and the element

at failure with principal stresses causing the failure.

(a) (b)

Fig. 6.4 (a) Mohr‟s failure criteria; (b) Element at failure showing principal stresses

So, if some how, the principal stresses at failure can be estimated, a Mohr‟s circle can be constructed

to represent this state of stress. Similarly, several tests to failure at various combination of principal

strsses would lead to construct several Mohr‟s circle. Such a series is plotted in Fig. 6.5. Since the

Mohr‟s circles are determined at failure, it is possible to construct limiting or failure envelope of the

shear stress. This envelope is called the Mohr‟s failure envelope which expresses the relationship

between f and f as represented by Eq.(6.2) and Fig. 6.4(a). It is to be taken in to note that the

failure envelope defined by Mohr (Fig. 6.4a) is a curved line.

It is worth mentioning that Mohr circle lying below the envelope, shown as a dotted circle in Fig.

6.5, represents a stable condition and the circle going above the failure envelope cannot exists.

Failure occures only when the combination of normal and shear stresses is such that the Mohr‟s

circle is tangent to the Mohr‟s failure envelope. If this envelope is unique for a given material, then

the point of tangency of the Mohr failure envelope defines the condition on the failure plane at failur.

Using the Pole method, the angle of friction plane from the point of tangency of the Mohr corcle and

the Mohr failure envelope. That is, Mohr,s failure hypothesis states that the point of tangency of the

Mohr failure envelope with the Mohr circle at failure determines the inclination of the failure plane.

Coulomb (1776) was interested in the sliding friction charactersistics of different materials and

observed that there is a normal stress independent component of the shear strength and a normal

stress dependent component. The normal stress dependent component is similar to sliding friction in

solid and he termed it as angle of internal friction and denoted by . The other component appears

to be related to the intrinsic cohesion of the material and denoted by the symbol c. Coulomb gives

the following linear equation for shear strength of soil involving parameters and c.

tancf (6.3)

Eq. (6.3) is presented in terms of total stress, on the failure plane. The equation can be rewritten

for effective stresses as follows.

tantan cucf (6.4)

where, c and are called effective stress parameters as against total stress parameters c and

respectively. u is the pore water pressure.

f

f 3

1

f = ( f)

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Fig. 6.5 Mohr circle at failure and Mohr‟s failure envelope

It is always easier to work with straight line relationship and the simpler thing is to do is to

straighten out the curved failure envelope of Mohr within the stress range encountered from

foundation soil. Accoirding to Coulomb it is sufficient to approximate the shear stress on the failure

plane as a linear function of the normal stress. Thus was born of Mohr Coulomb failure theory that

at failure the stress condition at the rupture surface can be defined by the straight line equation (Eq.

6.3 or 6.4) given by Coulomb touching the circle of stresses given by Mohr. It is a point of concern

that if the shear strength parameters c and are to be obtained from failure envelope, it is obvious

that the plot of and uses similar scales in both the co-ordinates.

6.5 Determination of Shear Strength Parameters c and

Shear strength (its parameters) of soils can be determined both in the laboratory or in the field. The

tests for studying the shear strength and related deformation characteristics of soils include,

Direct shear test

Triaxial test

o Triaxial Compression

o Triaxial Extension

Unconfined compression test

Vane shear test

Torsion shear test

Ring shear test (double shear)

Hollow cylinder test

Plain strain test

Cuboidal or true triaxial test

Borehole shear test

The first three are the most common laboratory tests. The fourth one, vane shear test, in

principle is similar both in the laboratory and field except that the apparatus used in the field is larger

in size for practical reasons. The vane shear test is usually done to determine the undrained shear

strength parameter (c) of saturated cohesive soil.

The other tests are mostly used for research purposes and hence are not dealt with here. There are

also some indirect ways of measuring the shear strength of soil in the field like Standard

Penetration Test (SPT), Cone Penetration Test (CPT) etc.. These tests involve the measurement

of resistance of the soil to penetration of a hollow cylinderical spilt tube or a cone while driven or

pushed in to the soil at a desired depth. These procedures are rather empirical in nature, though

widely been used through out the world for their simplicity and inexpensiveness. However,

engineering judgement and experience play the most important role to correlate the results which

depends on the type of the soil and also on the other factors. The common laboratory tests and their

salient features are described below.

Failure envelope

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Direct Shear Test

The direct shear test is a simple and widely used test for determing the shear strength of soil. The

direct shear apparatus is essentially a rectangular or circular box having separated lower and upper

halves (sections). A schematic diagram of direct shear apparatus is shown in Fig. 6.6. The lower half

of the box is fixed to a frame whereas, the upper section is capable of moving horizontally relative to

the lower one. The soil sample ois placed in the box, with approximately half of the sample within

either section. The sample size is usually 60 mm square or 75 mm circular having a thickness in

between 20 mm to 25 mm. In case of cohesive soil prism of soil is eithet prepared or taken from

the undisturbed sample using a cutter of similar dimension of shear box. Whereas, for cohesionless

soil, the specimen has to be prepared in the box at the required void ratio. Porous disc may be placed

on top and bottom of the specimen to facilitate the drainage condions. A normal load is applied to

the plane of shear via a loading plate placed over the sample (Fig. 6.6). The upper half of the shear

box is then moved laterally forcing the sample to shear across the plane between the two halves of

the box either by controlling rate of strain or stress.

In strain controlled apparatus, the shearing deformation is continuously applied at a constant

speed and shear force is measured by means of a proving ring or a load transducer. Such a test

can be continued even after the failure of the specimen. In stress controlled apparatus, the

magnitude of the shear stress is increased uniformly or in increments. In case of load

increment, each increment is applied and held constant until the shearing deformation ceases. A

stress contrilled test can not be continued as soon as the specimen fails and as such the strain

controlled preferred. Not only that a mechanically operated strain contrilled apparatus is simplest to

devise. Direct shear test

Shear box

Loading frame to

apply vertical load

Dial gauge to

measure vertical

displacement

Dial gauge to

measure horizontal

displacement

Proving ring

to measure

shear force

(a)

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Direct shear test

Leveling the top surface

of specimen

Preparation of a sand specimen

Specimen preparation

completed

Pressure plate

(b)

(C)

Fig. 6.6 Dircet shear apparatus (a) set up; (b) sample preparation; (c) Schematic diagram

In direct shear test the test procedure is repeated using atleast three identical specimen with variable

normal stresses.

It is sometimes convenient to interpret the results, if the normal stresses are so choosen that one is

closer to the existing effective overburden and the one of each of the other two on the either side of

the overburden. Usually, a record of magnitude of the shearing force and the shear displacement is

maintained to obtain the peak shear stress (failure stress) from the shear stress-strain plot for a

particular normal load. Changes in sample thickness that occurs during the shearing process are also

recorded so that the volume change versus shearing stress or shearing strain can be studied.

Test Results

Typical shear stress displacement and volume change displacement curves for a single specimen (one

normal load) are shown in Fig. 6.7. The slope of the stress strain curve is known as modulus of

elasticity or tangent modulus of elasticity of soil, ES. This modulus varies directly with the stiffness

of a soil; the more its value, the more is the stiffness and strength. The peak value of shear stress or

the maximum value at a relatively higher shear deformation (normally 20 per cent) gives shear

strength of the soil at a particular normal load.

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f1

Normal stress = 1

Direct shear tests on sands

How to determine strength parameters c and

Sh

ear

str

ess,

Shear displacement

f2

Normal stress = 2

f3

Normal stress = 3S

he

ar

str

ess

at

fail

ure

,f

Normal stress,

Mohr – Coulomb failure envelope

(a)

Direct shear tests on clays

Failure envelopes for clay from drained direct shear tests

Sh

ear

str

ess a

t fa

ilu

re,

f

Normal force,

Normally consolidated clay (c’ = 0)

In case of clay, horizontal displacement should be applied at a very

slow rate to allow dissipation of pore water pressure (therefore, one

test would take several days to finish)

Overconsolidated clay (c’ ≠ 0)

(b)

Fig. 6.7 Direct shear test results; (a) Cohesionless soil, (b) Cohesive soil (Drained Test)

A best possible straight line is fitted to the observed point to represent the failure envelope. The

shear strength parameters c and is thus obtained from the plot. It is to be noted that usually we get the drained strength parameters from direct shear test. For granular soil as the cohesion is zero

the available strength parameter would be only .

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The effective angle of friction for a particular density is obtained by plotting the maximum value of

shear stress f against the effective normal stress . Atleast three tests are carried out at different

normal stresses and a straight line passing through the origin and the respective points defines the

failure envelope for sand soil ( as c= 0). The slope of the line gives the value of . Typical values of

are shown in Table 6.1.

Table 6.1 Typical values of effective angle of internal friction for granular soil (after Terzaghi

and Peck, 1967)

Soil type Effective angle of internal friction, (degrees)

Loose Dense

Nonplastic silt 27-30 30-34

Silty sand 27-33 30-35

Uniform sand 28 34

Well gradred sand 33 45

Sandy gravel 35 50

Other observations

The design of normal shear box does not allow control of drainage of the sample. Sands and

gravels are free draining materials and they will shear under fully drained conditions. Clay deposits,

however, depending on the rate at which the soil mass is stressed, a soil element in the field may fail

either completely undrained (no dissipation of excess pore water pressure), partially drained (some

dissipation) or fully drained (complete dissipation). Although attempts could be made to measure the

undrained strength by shearing a sample rapidly in a few minutes, beacause there is no control over

the drainage there would be a degree of uncertainty as to whether this represented the true value of

the undrained strength. For this reason the undrained shear strength of a clay soil is usually measured

by the more sophisticated triaxial test.

The direct shear test may, however, be used to measure the drained strength of clay soils by first

consolidating the sample fully under the applied normal load and then shearing the sample at a

sufficiently slow rate to ensure the fuul dissipation of pore water pressure generated by the applied

shear stress. Hence, for drained clays and sands the effective stress normal to the shear plane is given

by = N/A and the shear stress = S/A, where N and S are applied normal and shear force

repectively on the plan area of shear box, A.

Typical stress strain volume change relationships for drained tests on loose and dense sand are

shown in Fig. 6.8. The shearing in a loose sand rearranges the particles to fill into the voids causing a

denser packing and a decrease in volume. At larger strains of (approx. 20%) the sample shears at a

constant volume under a constant value of shear stress. At this large strains the tendency for a

volume increase as some particles move up and over each other is foiled by adjacent particles

moving down into the void space created, resulting in zero net volume change. This condition of soil

is known as critical void ratio state. For a dense sand, the interlocking grains have to move apart in

order to allow relative movement or shear between the particles to occur. Thus the sample expands

during shear and this phenomenon is known as dilatancy. The consequent result is the work against

the confing pressure.The peak shear stress therefore occurs at that particular value of shear strain at

which dilatancy rate is maximum. With increasing shear strain, the dilatancy rate decreases as the

sample approaches to a constant void ratio and shear stress decreases to a residual value. For the

same confining pressure, the residual shear stress of a dense sample is equal to the maximum shear

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stress of a loose sample. Typical failure strains for loose sands are around 12-16 percent and

for dense sands 2-4 percent.

Stress-strain-volume change relations for normally consolidated and over consolidated soils (for

drained condition) are very much similar to loose and dense sands respectively. However, the

normally consolidated clay will have only friction parameter( ) which infact dictates its

depositional behaviour. Overconsolidated clay test results would be in the form of c and . The

value for clay soil is not influenced appreciably due to over consolidation and generally

ranges from about 20o to 30

o, decreasing with increasing plasticity index. Overconsolidated

clays are characterised by an intercept on axis which is the effective cohesion c . Its value

usually ranges from 5 to 30 kN/m2. For both normally consolidated and over consolidated soils

there is a decrease in shear stress from its peak to residual, and for the latter its is prominent. The

residual angle of friction may be even 9o less than that at peak stress. The stress strain volume

change behaviour of various types of soil are shown in Fig. 6.8 Direct shear tests on sands

Sh

ea

r s

tre

ss

,

Shear displacement

Dense sand/

OC clay

fLoose sand/

NC clayf

Dense sand/OC Clay

Loose sand/NC Clay

Ch

an

ge

in

he

igh

t

of

the

sam

ple Exp

an

sio

nC

om

pre

ssio

n Shear displacement

Stress-strain relationship

Fig. 6.8 Stress-strain-volume change relationships for drained tests on clays

Triaxial Test

The triaxial test is perhaps the most versatile and sophisticated test for determining the shear

strength of soils. Triaxial tests are basically of two types depending on the mode of application of

load to represent the field conditions; (i) triaxial compression test and (ii) triaxial extension test.

However, it is very unsual where the latter type of test condition prevails. As such, type (i) test is the

most conventional one and meets most of the purposes, is briefly described herein.

In the conventional triaxial compression test, a cylindrical soil sample usually 38 mm in diameter

and 76 mm in length, is prepared and encased in a thin rubber membrane which extends over a top

cap and bottom pedestal. It is then placed in a cylindrical perspex chamber where an all around or

confining pressure can be applied. The specimen of other dimensions can also be tested provided the

the length diameter ratio should lie in between 2:1 to 3:1. To facilitate proper drainage condition

of the sample, porous stones/discs and filter papers are used at the edges of the sample. In order to

enhance further drainage, filter paper strips are wrapped around the sample. These connect the

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porous disc at the top cap, the take off being through a nylon tube which passes out of the cell

through its base. A schematic diagram of triaxial test set up is shown in Fig. 6.9.

Fig. 6.9 A schematic diagram of triaxial shear test set up.

The prepared soil sample is properly placed in the cylinder, the cell is the filled with water via cell

pressure control tube and subjected to a desired cell pressure. The water pressure, commonly

referred to as confining pressure, acts horizontally on the cylindrical surface of the sample through

the rubber membrane as well as vertically through the top cap. The pore water pressure in the

sample is measured through a porous disc at the base pedestal and connected through a water filled

duct to a pressure transducer. To minimise the friction at the top and bottom of tha sample and

allow an unrestricted lateral deformation during shear, greased rubber discs are placed between the

sample and the end caps.

An axial load commonly referred to as deviator load (stress), is the applied and steadily increased until the failure of the specimen occurs. The procedure is repeated for different confining pressures and corresponding deviator loads atleast three times using three identical soil specimen. A Mohr‟s

circle for each combination of confining pressure ( 3), and total axial stress ( 3 + deviator stress =

1) is constructed. The tangent to the resulting circles becomes the Mohr envelope as shown in Fig. 6.10.

A representation of stressed specimen is shown in Fig 6.10(a). In Fig 6.10(b), 31, 32 and 33

represents the confining pressure ( 3) for specimen 1, 2 and 3 respectively; the corresponding axial

stresses are respectively 11, 12 and 13.

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(a) (b)

Fig. 6.10 Results of triaxial test; (a) Representation of stressesd specimen; (b) Mohr‟s circle

and failure envelope for three differenr combinations of principal stresses.

In unavoidable circumstances, a smaller number of specimen can be used. Some times a single specimen can be loaded to near failure for a particular confining pressure, then the confining pressure is increased to next higher stage and again loaded near to a failure, and then the final cell pressure is applied to loaded it to failure. This type of test is commonly referred to as stage loading. However, in all these circumstances the interpretation of results calls for careful engineering judgement skill. The lateral pressure produces a normal horizontal stress on all vertical planes of a specimen. Hence

any two vertical planes of the specimen experience equal stress 3 (Fig. 6.10.a). No shear stresses are produced on the horizontal planes or any two orthogonal vertical planes by either the axial load or the confining pressure. That is, the vertical and horizontal planes of the element shown are

subjected only to maximum and minimum normal stresses 1 and 3 (principal stresses), respectively. The triaxial test is sohpisticated in the sense that it may be manipulated to within a reasonable estimated insitu conditions. The apparatus is designed to allow;

Control of confining pressure and deviator stress

Control of porwater pressure and drained water

Saturation of soil sample

As such a wide variety of tests are possible in triaxial shear testing to depict insitu loading and drainage conditions, as shown in Fig. 6.11.

1

1

3 3

Failure envelope

31 32 33 11

12 13

c

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- 16 -

Types of Triaxial Tests

Is the drainage valve open?

yes no

Consolidated

sampleUnconsolidated

sample

Under all-around cell pressure c

Step 1

Is the drainage valve open?

yes no

Drained

loading

Undrained

loading

Shearing (loading)

Step 2

CD test

CU test

UU test

Fig. 6.11 Types of triaxial test

The following tests are adequate for most of the engineering purposes.

Unconsolidated Undrained test (UU or Quick or Q test)

Consolidated Undrained test (CU or Qc or R test*)

Consolidated Drained Test (CD or Slow or S test)

(* it is simply because, R lies between Q and S in alphbatic order as is the case of CU with respect

to UU and CD)

Generally, there are three different and successive steps in a triaxial test. These are:

Preparation of soil sample

Application of cell pressure

Application of excess axial stress at constant cell pressure

The first step, preparation of soil sample, for all the soil sample is essentially similar with only

a minor exception that in case of UU test the drainage accessories like wrapping of filter paper

around the soil specimen can be ommitted. The dissipation of pore water pressure or drainage during second and third steps of testing depends upon the test type. Often pore water pressure is measures in consolidated undrained test to estimate the effective stress parameters. Such a test is

designated as UC test. Table. 6.2 presents the drainage and loading conditions during second and

third steps.

Table 6.2 Drainage and loading conditions in different types of triaxial tests.

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- 17 -

Type of test Second Step: Application of

cell pressure

Third Step: Application of additional

axial stress at a constant cell pressure

Unconsolidated

Undrained (UU)

Drainage is not allowed.

Sample is unconsolidated.

Drainage is not allowed. That is, failure

occures in undrained condition.

Consolidated

Undrained (CU)

Drainage is allowed. That is,

sample is consolidated.

Drainage is not allowed. That is, failure

occures in undrained condition.

Consolidated Drained

(CD)

Drainage is allowed. That is,

sample is consolidated.

Drainage is allowed. That is, failure occures

in drained condition.

Shear strength parameters obtained from various types of triaxial test

The triaxial test data obtained from a test on each of the specimen are separetely plotted as devator

stress strain diagram to obtain the peak or maximum deviator stress and the corresponding axial

strain. The excess pore water pressure reading at the failure strain is used to calculate the effective

principal stresses where appropriate. Thus the principal stresses on the soil sample are estimated and

Mohr circle are drawn. For details of the test procedure and computations reference can be made to

Lambe(1993).

Shear strength parameters for cohesive soil

The shear strength of cohesive soils depends on many factors like degree of saturation, stress

history, loading rate and drainge conditions. Cohesive soils are often treated in a saturated state

to represent the worst condition that can prevail in the field, though the parameters estimated wil be

generally on the conservative side. Even then, researchers, concentrated many of their studies on

unsaturated and remoulded clays to assess the insitu behaviour by analysing the results obtained

from their studies.

Unconsolidated undrained (UU) test

In this test a minimum of three soil samples are subjected to different confining pressure 3 and

then loaded to failure. Before applying this confining pressure, a small confining pressure and equal

amount of pore water pressure may be applied to the soil and allowed to stay for some time to

saturate the sample. This application of pore water pressure to the sample is known as back

pressure. The degree of saturation of the soil specimen can be checked from the pore pressue

parameter, B as discussed later.

When a saturated cohesive soil is subjected to a stress increase, it will respond simply by increasing

the pore water pressure by that amount. So, any change in all around pressure or axial stress will not

bring any change in the structure or arrangement of soil particles within the soil mass. Hence the

undrained strength remains constant and independent of cell pressure. That is, the soil will fail at an

equal deviator stress for all the different cell pressures. Deviator stess which is the diameter of the

Mohr‟s circle being equal, the failure envelope will be a horizontal line thus depicting that only

cohesion parameter will contribute to generate the shear stength of soil. The cohesion parameter at

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- 18 -

this condition is termed as undrained cohesion and designated as cu. The other parameter would

be zero. A typical example of the Mohr,s circle, failure envelope and the estimated shear parameter

is shown in Fig. 6.12. The failure envelope in this case can be defined as

uf c (6.5)

Fig. 6.12 Unconsolidated undrained (UU) strength of saturated clay

Consolidated undrained (CU) test

In this test soil sample is initially consolidated under confining pressure. This pressure is known

as initial effective confining pressure. The volume change reading can also be taken at this stage to

estimate the initial void ratio of the specimen. After consolidation, the confining pressure and back

pressure may be increased by equal amount thus keeping the effective cell (confining) pressure

constant. As usual, the back pressure is applied to saturate the soil sample and also to facilitate the

measurement of negative pore water pressure during application of deviator stress in case of an

overconsolidated clay. The soil sample is sheared at this elevated cell pressure ( 3) with a deviator

stress of , that is axial stress of 1(= 3+ ). Morh‟s circles are drawn for the three specimens to

determine the failure envelope and strength parameters in terms of total stress. The Mohr‟s circle

diagram and failure envelope will be similar to the generalised diagram of Fig. 6.10b as reproduced

in Fig. 6.13. The strength parameters would be designated as ccu and cu instead of c and as

mentioned there. The failure envelope will be represented by

cucuf c tan (6.6)

As mentioned earlier, consolidated undrained test can also be carried out with pore water pressure

measurement while shearing the soil sample. The effective stress parameters in such a case are

determined drawing Mohr circle and hence the failure envelope taking in to consideration the

effective cell pressure 3 as ( 3-uf) and effective axial stress 1 as ( 1- uf). Here uf is the pore water

pressure at failure. Thus the effective minor and major principal stresses 1 and 3 respectively are

to be computed for all the three specimens.

cu

Failure Envelope

31 32 11 33

12

13

u=0

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CU tests How to determine strength parameters c and

Devia

tor

str

ess,

d

Axial strain

Sh

ea

r s

tre

ss

,

or ’

d)fb

Confining stress = 3b

3b 1b3a 1a

( d)fa

cuMohr – Coulomb

failure envelope in

terms of total stresses

ccu

1 = 3 + ( d)f

3

Total stresses at failured)fa

Confining stress = 3a

Fig. 6.13 Consolidated undrained (UU) strength of saturated clay

It is understandable that the Mohr circles for this effective stress condition would be of similar diameter as those for total stress condition. Only thing is that they will shift towards left in the

diagram thus resulting in a higher and lower c values interms of effective stress A typical illustration is shown in Fig 6.14.

Fig. 6.14 Total and effective stress parameters from a CU test (Typical one circle of each are

shown, others are hidden)

The effective strength parameters are ccu and cu respectively for cohesion and friction and the

equation for failure envelope is

cucuf c tan (6.7)

However, for normally consolidated soil ccu would be zero as explained later.

cu

cu

ccu

ccu

uf

Total stress

circle

Effective

stress circle

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Consolidated drained (CD) test

In this type of test, the sample is first consolidated under confining or cell pressure. Back pressure may be applied in a similar manner to that of CU test to saturate the sample, if necessary. Thus the

confining pressure 3 here is also the effective confining pressure 3 as that of CU test. The specimen is then sheared to failure with a very small strain rate so that no excess pore water pressure can develop on the failure surface. That is, the pore water pressures remains zero at all the stages of the test thus prevailing effective stress conditions althrough. The volume change reading can also be taken to examine the void ratio of the soil at failure condition. Mohr‟s circle interms of effective minor and major principal stresses at failure for all the three specimen are are drawn to obtain the

effective shear strength parameters c and . The failure envelope is of typical in nature as

represented by Fig. 6.10b, and c and would be replaced by the parameters c and . Fig. 6.15 shows the related diagrams. The values of these parameter are different. The failure envelope can be given by:

tancf (6.8)

The parameters c and are often replaced by cCD and CD to remind that these parameters are for drained condition. The parameters obtained from a drained test always refers to effective stress conditions though these can also be obtained from a CU test and that should be specifically spelled out. However, in laboratory testing there may be slight differences in effective stress parameters determined from CU test and CD tests. However, the magnitude strength obtained from a CD test is always higher than that form a CU test for a normally consolidated clay.

(a) (b)

Fig. 6.15 Shear strength parameters from a consolidated drained (CD) test.

Strength parameter for cohesionless soil

When load is applied to a soil, the soil is stressed and causes the soil-water matrix to compress and

some of the water is squeezed out. Because cohesionless soils have a high hydraulic conductivity

(coefficient of permeability), this water is able to move quickly and easily. The potential drainage

rate is atleast as high as the loading rate. This condition is known as drained condition and the

stresses acting on a cohesionless soil are always the effective stresses. Hence, in case of cohesionless

soils the measured strength parameters from shear tests are always the effective one and it is almost

′1

′1

3

′3

Failure envelope

′31 ′32 ′33 ′11

′12 ′13

c′

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- 21 -

independent of type of triaxiat tests. As cohesionless soil has smaller surface forces, the cohesion

parameter can be considered as zero the the strength envelope can be explained by the equation

tanf (6.9)

As such, only one Mohr‟s circle is required to define the straight line failure envelope of strength

as another point is known to be the origin of - coordinate system. However, confining pressure

has a little effect on strength the parameter and as such, conventionally three specimen are tested at

different confining pressures, Mohr‟s circles are drawn to obtain the failure envelope of the form

represented by Eq. (6.9) and thus the angle of internal friction is determined. A typical Mohr

circle diagram for triaxial test on dry cohesionless soil is shown in Fig. 6.16, where the strength

parameter c is zero.

Fig. 6.16 Shear strength results of cohesionless soil from a triaxial test

However, the moist cohesionless soil might show a very small value of apparent cohesion which is

due to surface tension. In all the shear strength test of sand volume change is a very important

factor. Hence, it is preferable to record the volume change readings for better explantion of the test

results.

It is to be noted that due to the mode of formation of failure planes, the shear strength

parameter obtained from direct shear test overestimates the triaxial value approximately

by 10%.

Strength of normally consolidated and over consolidated clays

Clay may exist in the field under two conditions;

existing effective overburden pressure, o is greater than or equal to the maximum of

effective pressures subjected in the past called overconsolidation pressure oc. This type

of soil is known as normally consolidated soil.

existing effective overburden pressure ( o) is less than the maximum past

pressure( oc). This type is over consolidated soil.

The ratio oc/ o is known as over consolidation ratio (OCR). Strength characteristics of normally

consolidated and over consolidated clays (saturated) differ because of the fact that their depositional

modes and drainage behaviour are responsible to impart the cohesion and frictional properties

between the particles. As such the strength characteristics of clay soils are dependent on OCR.

Failure envelope

′31 ′32 ′33 ′11

′12 ′13

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- 22 -

The UU test on both normally consolidated and overconsolidated clays gives c parameter that is

shear strength in this case equals to c only. However, the strength for a overconsolidated clay is

higher than the similar clay in a normally consolidated state. The difference may be attributed to the

increased density and the negative pore pressures in over consolidated states. The strength measured

in a CU test comes from both c and parameters in each of the caes. While compared the strength,

it depends on the over burden pressure and OCR.

For a CU test with pore pressure measurement or CD test , normally consolidated soil gives

(drained ) parameter resulting in a zero cohesion intercept. However, for over consolidated soils

we get both c and . Strictly skeaking, the failure envelopes for overconsolidated clays are

nonlinear. For a over consolidated soil, if the failure envelope of the normally consolidated range is

extended backwards it will pass through the origin giving c = 0.

Undrained shear strength of normally consolidated clays varies with effective overburden pressure

(i.e. with depth). Several empirical relations are available. Of them, the one given by Skempton is

mentioned below.

p

o

u Ic

0037.011.0 (6.10)

Where, cu is the undrained cohesion, o is the effective over burden pressure and IP is the plasticity

index in percent. Eq. 6.10 afford a quick but approximate estimation of the increase in undrained

shear strength of a normally consolidated clay with depth.

Unconfined Compression Test

Unconfined compression test is essentially a special case of triaxial test where the minor pricipal

stress or cell pressure is kept as zero. No rubber membrane is required to encase the soil sample As

such, the apparatus becomes much simpler. This is the easiest, quickest and simplest test for

determining the undrained strength of saturated cohesive soil. This test can also be done on

unsaturated sample, but it should be borne in mind that undrained shear strength of cohesive soil

varies significantly with the moisture content and density and in the design due consideration should

be given. Similar to triaxial tests, three cylindrical specimens are tested for axial compression to

failure in an unconfined compression machine and their average values are taken as the strength.

The unconfined compression testing macine essentially consists of a loading frame where the

cylindrical soil sample can be placed in between base platform and a top cap. The top cap is attached

to either with a proving ring or load transducer. The bottom platform can be raised either by gear

system or by a hydraulic pump system. A view of the machine is shown in Fig. 6.17 and the failure

envelope is shown in Fig. 6.18.

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- 23 -

Fig. 6.17 A view of unconfined compression testing machine

Theoretically, an unconfined compression test should yield the same results of undrained shear

strength as obtained from triaxial UU test. However, this supposition is only valid if several

asumptions are satisfied.

The soil specimen must be clay and 100 percent saturated.

The specimen must not contain any fissures, cracks, silt seams, verves or any other

defects.

The specimen must be sheared rapidly to failure. If the time is too long, evaporation and

surface drying may result in high strength. Typical time to failure is 5 to 15 minutes. If

the sample does not show any sign of failure, the unconfined comprresive value is taken

as the as 20 percent strain.

Fig. 6.18 Mohr circle diagram for unconfined compression test

Unconfined compressive strength also gives a measure of consistency of clay soil as expressed in

Table 6.3. Unconfined compression test can be done both on undisturbed and remoulded samples.

Their values usually gives some useful estimation about the sensitivity of clay soils.

cu = qu /2

Failure Envelope

3=0 1=qu

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- 24 -

Table 6.3 General relationship of consistency and unconfined compressive strength of

clay soil.

Consistency Unconfined compression strength,

qu(kN/m2)

Very soft < 25

Soft 25 - 50

Medium 50 - 100

Stiff 100 - 200

Very stiff 200 - 400

Hard 400 - 800

Sensitivity describes the effect of stress and environment with time on the strength of the soil. It is

usually defined as the ratio of undrained shear strength of clay soil in undistubed condition to that at

a remoulded state at the same water content. Though, it is referred to undrained strength it is

customary to use the unconfined compressive strength to define its sensitivity. Hence,

)(Re

)(,

mouldedu

dundisturbeu

tq

qSySensitivit

The sensitivity of a soil measures the reduction of shear strength that might occur due to disturbace

of the sample. Bjerrum (1954) gives a sensitivity classification os soil which is presented in Table

6.4.

Highly overconsolidated soils are insensitive. This is partly due to low natural water content in the

soil deposit. Glacial clays lie in the range of medium to extra sensitive classification. Some of fresh

water and marine deposits are quick. However, sensitivity most of the clay deposits ranges from 2 to

4. The disturbance or the decrease in strength of clay soil is attributed atleast partly due to

disturbance of adsorbed water in clay layers. If the remoulded soil is left undisturbed at the same

water content, partly of its strength is regained due to reorientation of clay particles. This is known

as thixotropy.

Table 6.4 Sensitivity Classification of Clay Soil

Sensitivity Value Soil Description

Less than 2 Insensitive

2 -4 Medium sensitive

4 - 8 Sensitive

8 - 16 Very sensitive or Extra Sensitive

16 - 32 Slightly quick

32 - 64 Medium quick

Greater than 64 quick

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Vane Shear Test

The vane shear test can be applied for measurement of undrained shear strength of clay soil both in

the laboratory and field. In particular this test is very suitable for soft clays, which otherwise may

be greatly disturbed during the extraction and testing process.

Essentially, the vane shear apparatus consists of a series of thin rectangular blades welded to a

circular shaft, so that when the shaft is rorated about its axis each blade generates the same cylinder.

In its usual form, four blades are used forming a cross section. A schematic diagram of vane shear

apparatus is shown in Fig. 6.17(a).

Fig. 6.17 Vane shear test; (a) diagram of apparatus; (b) distribution of stresses

The vane dimensions may vary but a height/diameter ratio of 2.0 is commonly used. The vane is

pushed into the soil to a predetermined depth and rotated rapidly (usually 0.1o/sec) until shear takes

place over the surface area approximately equal to that generated by the vane. The strength is

estimated from the maximum measured torque required to shear the cylindrical surface. A simple

formula between shear strength and torque can be derived depending on the following assumptions.

The soil is purely cohesive ( = 0), homogeneous and isotropic.

The insertion of the blade causes no disturbance to the soil.

No drainage takes place during shearing.

Failure takes place by shearing over the surface of the cylinder generated by the

rotating vane.

No progressive failure takes place in the soil and the shear strength is fully mobilsed on

the surface of the cylinder at failure.

The total torque is the sum of the opposing moments on the cylinder side and on the two edges of

the cylinder. Mathematically, it can be written as:

EdgeSide MMT 2

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- 26 -

Where,

T = total applied torque

Mside = resisting moment due to the shear forces on the cylindrical surface

Medge = resisting moment due to shear forces on a single edge surface.

Let us consider a cylinder of soil of height, h and diameter, d (Fig. 6.19) will resist the torque until the soil fails. If the assumption of full shear mobilisation at the edges are not valid, there could be various assumptions regarding the distribution of mobilised shear strength at the edges. The possible distributions are shown in Fig. 6.19(b). Accordingly, the expression for resisting torque can be obtained by integrating the resulting moment for the relevant stress distribution diagram of the edges and adding the value for the moment due to surface resistance. The general expression can be obtained as follows:

4

ad

2

hdcT

32

u (6.13)

where, cu = undrained shear strength of soil

a = 32

for uniform distribution of shear at the edges ( fully mobilised strength).

= 21

for triangular distribution of shear at the edges.

= 53

for parabolic distribution of shear at the edges.

Sometimes, a vane shear test is done in a borehole where to top edge of the vane concides with the bottom of the borehole. In such a case only one edge instead of two partakes in the resistance process. The value of „a‟ in Eq. (6.13) will then be reduced to half of the values as given above. For a conventional event, where a uniform distribution at both the edges is considered and the height of the vane is twice the diameter of the failure cylinder, Eq. (6.13) takes a simplified form to estimate the undrained cohesion of soil as:

37

6

d

Tcu (6.14)

6.6 Pore Water Pressure Parameters

Pore water pressure plays a significant role in determining the shear strength of soil. The change in

pore water pressure due to change in applied stress is characterised by dimensionless coefficients

called ‘pore pressure coefficients’ or ‘pore pressure parameters A and B’. These parameters have

been proposed by Skempton(1954). In undrained triaxial compression test, pore water pressure

develops during the application of cell pressure and also during the application of additional axial

stress.

The triaxial test, in addition to determining the strength parameters of soil, also gives data for

predicting the excess the excess pore water build up at a point in a soil mass by a change in total

stress condition.

The ratio of pore water pressure developed due to application of confining or cell pressure is termed

as B parameter: Therefore,

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- 27 -

3

c

c

cuu

B (6.15a)

That is, uc = B 3 (6.15b)

where, c = increase in confining pressure

uc = increase in pore pressure due to c

3= increase in minor principal stress.

In undrained condition, the decrease of volume of mass is equal to that in the volume of pores.

Using the principle of theory of elasticity it can be shown that

sk

vo

C

nCB

1

1 (6.16)

where,

n = porosity of soil

Cvo = volume compressibility (change in volume per unit volume per unit pressure

increase) of the voids

Csk = volume compressibility of soil skeleton.

For delails of derivation reference can be made to Holtz and Kovacs(1981).

In case of a saturated soil, the volume compressibility of void (Cvo) equals to volume compressibility

of water, Cw. The compressibility of water as compared to that of soil skeleton is very small and as

such

sk

vo

C

C can be considered as zero for all practical purposes. Hence, for a saturated soil mass,

accoring to Eq. (6.15), the parameter B equals to unity.

On the contrary, for a dry soil Cvo takes the value of volume comprssibility of air, Ca which is much

greater than Csk. Hence,

sk

vo

C

Cbecomes to infinity resulting in a zero value of B. The degree of

saturation dictates the value of B in the other instances. The variation of experimentally determined

values of B with respect to degree of saturation are presented in Fig. 6.18.

Fig. 6.18 Variation of pore pressure parameter B with degree of saturation, Sr of clay soil

(after Venkatramaiah, 1995)

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- 28 -

The value of B depends also on soil type, stress level and stress history. The exact relationship for a

particular soil will have to be determined experimentally. Pore water pressure developed during the

application of deviator stress in a triaxial machine can be related as

31

du

A

That is, )(31

Aud

(2.17)

where, ud = increase in pore water pressure due to increase in deviator stress ( 1- 3).

A is a dimensionless experimentally determined parameter which is the product of another

parameter, A and B. If now u denotes the total excess pore water pressure within the soil mass due

to increase in confining pressure and deviator stress, then using the principle of superposition, it can

be related as:

)(.

)(.

313

313

ABuB

AuBuuudc

That is, u = B u3 + A ( 1- 3) (6.18)

For a 100% saturated sample, where B = 1 the equation simplifies to the form:

u = u3 + A ( 1- 3) (6.19)

Similar to parameter B, the parameter A is not also constant. It varies with the soil type, stress level,

soil anistropy, density index and stress history (OCR). Bishop and Hankel(1962) gave a relationship

between over consolidation ratio and A value at failure, Af. This is shown in Fig. 6.19. Typical Af

values for different clay soils are also given by Skempton. They are presented in Table 6.5.

Skempton‟s pore pressure coefficients A and B are the most useful parametersin engineering practice

in judging the shear strength of soil since they can be used to predict the induced pore pressure

whenever the changes in the total stress are known. For example, if the pore pressure response

during embankment construction on a very soft clay foundation (undrained loading) is to be

estimated, these parameters can be used. The increase in pore water pressure results in the indication

of instability. The factor of safety thus can be calculated at a particular stage of loading.

Fig. 6.19 Variation of Af with over consolidation ratio (OCR)

Table 6.5 Values of Af for various clay soils.

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- 29 -

Type of soil Value of Af

Highly sensitive clay 0.75 to 1.50

Normally consolidated clay 0.50 to 1.00

Compacted sandy clay 0.25 to 0.75

Lightly overconsolidated clay 0.00 to 0.50

Compacted gravelly clay -0.25 to 0.25

Heavily overconsolidated clay -0.50 to 0.00

Measurement of pore pressure parameters

The pore pressure parameters A and B may be determined in a consolidated undrained test. The

coefficient B is determined during consolidation stage, and A during the shear stage of the test.

All around pressure (cell pressure) is applied in the sample and drainage valve is kept closed. The

change in pore pressure uc is measured against the applied confining pressure 3. The parameter

B is obtained as

3

cu

B .

The drainage path is then opened and the sample is allowed to consolidate at the same confining

pressure. Once consolidated, the drainage valve is closed and a deviator stress ( 1- 3) is applied

keeping the cell pressure constant. The pore water pressure change, d

u at this stage is recorded.

The parameter is obtained as

31

du

A hence A as B

A.

6.6 Shear Strength of Unsaturated Soil

Shear strength of partially saturated soils

In the previous sections, we were discussing the shear strength of

saturated soils. However, in most of the cases, we will encounter

unsaturated soils

Solid

Water

Saturated soils

Pore water

pressure, u

Effective

stress, ’Solid

Unsaturated soils

Pore water

pressure, uw

Effective

stress, ’

Water

AirPore air

pressure, ua

Pore water pressure can be negative in unsaturated soils

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- 30 -

Shear strength of partially saturated soils

Bishop (1959) proposed shear strength equation for unsaturated soils as

follows

'tan)()(' waanf uuuc

Where,

n – ua = Net normal stress

ua – uw = Matric suction

= a parameter depending on the degree of saturation

( = 1 for fully saturated soils and 0 for dry soils)

Fredlund et al (1978) modified the above relationship as follows

b

waanf uuuc tan)('tan)('

Where,

tan b = Rate of increase of shear strength with matric suction

Shear strength of partially saturated soils

b

waanf uuuc tan)('tan)('

Same as saturated soils Apparent cohesion

due to matric suction

Therefore, strength of unsaturated soils is much higher than the strength

of saturated soils due to matric suction

- ua

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- 31 -

- ua

How it become possible

build a sand castle

b

waanf uuuc tan)('tan)('

Same as saturated soils Apparent cohesion

due to matric suction

Apparent

cohesion

Factors Affecting Angle of Shearing Resistance

The shear resistance of sands and gravels is made up of sliding and rolling friction plus the resistance

to volume change by interlocking. Interlocking effect is important for dense sands and is the main

contributing factor for the increased angle of shearing resistance in them. However, the angle of

shearing resistance corresponding to the ultimate strength is found to be not much different for the

dense sand and the loose sand. The angle of shearing resistance decreases rapidly with increase in

percentage of fine. A fine content of about 20 to 25 per cent may reduce the φ′ value for clean sand

by 5° to 7°. Table 6.5 shows the typical values for of φ′.

Table 6.5 Approximate values of φ′ for sand

Type of sand Round grains, uniform grain

size distribution

Angular grains, well graded

Loose sand 28.5o 34

o

Dense sand 35o 46

o

A rather elegant empirical formula suggested by Brinch Hansen and Lundgren (1960) takes into account

the influence of the major factors in determining φ′ in sands and gravels. It is expressed in the form:

ooooo

432136 (6.20)

In Eq. 6.20, φ′ = 36° represents the φ′ value for average conditions while φ′1, φ′2, φ′3 and φ′4

account for influence of grain shape, size, gradation and density index respectively.

Grain shape factor, φ′1, is taken on the following basis:

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Angular grain = + 1o

Subangular grain = 0o

Rounded grain = - 3o

Well-rounded grain = - 6o

Grain size factor, φ′2

Sand = 0o

Fine gravel = + 1o

Medium and coarse gravel = + 2o

Gradation factor, φ′3

Poorly graded soil = - 3o

Medium uniformity = 0o

Well graded soil = + 3o

Relative density factor, φ′4

Loosest packing = - 6o

Medium density = 0o

Densest packing = - 6o

The influence of relative density is the most important, as can be seen above. Values of φ′ can range

from 20o to 48

o.